Answer to Question #276285 in Physics for kei

Question

In a projectile laboratory experiment on, a mini-launcher was tilted 60.0° above the horizontal. A 10.0 g ball was placed inside the mini launcher compressing the spring inside the mini-launcher, whose force constant is 21.0 N/m, by a distance 𝑥. When the spring is released, the spring pushes the ball out of mini launcher and reaches a height of 67.5 cm above the muzzle of the mini launcher. Assume that the inner walls of the mini-launcher is frictionless.

(a) What is the speed of the ball as it leaves the muzzle of mini-launcher?

(b) If the distance travelled by the ball inside the mini-launcher from the compressed spring is 20.0cm, to what distance 𝑥 was the spring initially compressed?

(c) What is the distance 𝑑 between the muzzle of the mini-launcher and the uncompressed length of the spring?

(d) What is the speed of the ball when the spring returns to its normal length?

Answer 1

$$0.675=\frac{v^2\sin^2{60}}{2(9.8)}\\v=4.2\frac{m}{s}$$

$$21(x^2)=0.01(4.2)^2\\x=0.091\ m\\d=0.291\ m\\\\v=4.2\frac{m}{s}$$